Are there overlaps between infinite discontinuities and oscillatory discontinuities, such as piecewise function x < 0, f (x) = 1 / x, x > 0, f (x) = sin Is there any overlap between infinite discontinuity and oscillation discontinuity For example, when the piecewise function x < 0, f (x) = 1 / x, x > 0, f (x) = sin (1 / x), what is the breakpoint of x = 0?

Don't you understand the meaning of infinite discontinuities and oscillatory discontinuities?

Is the discontinuity of the function y = SiN x sin1 / x the third kind of discontinuity?

When x = 0, y is undefined, but there is a limit at x = 0
So:
The discontinuity of y = SiN x sin1 / X is x = 0,
Is the first kind of breakpoint (breakpoint can be removed)

The minimum positive period of the function f (x) = Sin & # 178; (2x - (π / 4)) is

f(x)=sin²﹙2x-(π/4))
={1-cos[2(2x-π/4)]}/2
=[1-cos(4x-π/2)]/2
=(1-sin4x)/2
So t = 2 π / 4 = π / 2

The minimum positive period of the function f (x) = Sin & # 178; (2x - π / 4) is
How does f (x) = 1 / 2 (1-sin4x) come from?

cos(2α)=1-2*sin²α
sin²α=(1-cos(2α))/2
So the minimum positive period of F (x) = (1-cos (4x - π / 2)) / 2 = (1-sin (4x)) / 2 is π / 2

How to find the maximum and minimum of F (x) = (sinxcosx) / (1 + SiNx + cosx)?

Let 1 + SiNx + cosx = t, that is, SiNx + cosx = T-1, the square of both sides get: 1 + 2sinxcosx = T ^ 2-2t + 1, that is, sinxcosx = (T ^ 2-2t) / 2. Bring in the original formula to get: F (x) = (T ^ 2-2t) / 2T = 2t-2

What is the range of y = (2-cosx) / SiNx
Ask for advice

Y = (2-cosx) / sinx-y = (cosx-2) / (sinx-0) - y is the slope of the line between the moving point on the unit circle and the fixed point (0,2), so - y ≤ - sqrt (3) or - y ≥ sqrt (3), that is, y ≤ - sqrt (3) or Y ≥ sqrt (3), so the range of y = (2-cosx) / SiNx is {y | y ≤ - sqrt (3) or Y ≥ sqrt (3)}

Find the maximum value of the function y = (2 + SiNx) (2 + cosx)

y=(2+sinx)(2+cosx)
=4+2(sinx+cosx)+sinxcosx
=[(sinx+cosx)²-1]/2+2(sinx+cosx)+4
=(sinx+cosx)²/2+2(sinx+cosx)+7/2
=(1/2)[(sinx+cosx)+2]²+3/2
Because SiNx + cosx = √ 2Sin (x + π / 4)
-√2≤sinx+cosx≤√2
therefore
When SiNx + cosx = - √ 2
The minimum value of Y is (1 / 2) (4 + 2-4 √ 2) + 3 / 2 = 9 / 2-2 √ 2
When SiNx + cosx = √ 2
The maximum value of Y is (1 / 2) (4 + 2 + 4 √ 2) + 3 / 2 = 9 / 2 + 2 √ 2

If f (x) = SiNx / (x ^ 2 + cosx) + √ 3, X is between - π and π, the maximum value is m and the minimum value is n, then M + n = 2 √ 3,

SiNx is an odd function
X ^ 2 + cosx is an even function
The function g (x) = SiNx / (x ^ 2 + cosx) is odd, and the sum of the maximum and minimum of G (x) is 0
M = the maximum value of G (x) + 3
The minimum value of n = g (x) + 3
M+N=0+√3+√3=2√3

The maximum value of cosx + SiNx + 2012 / cosx + 2012 is m, the minimum value is m, find m + M

cosx+sinx+2012/cosx+2012=1+（sinx/cosx+2012）
Let y = SiNx / cosx + 2012, then - radical (1 + Y & # 178;) < 2012y = SiNx ycosx ≤ radical (1 + Y & # 178;),
2012²y²≤（1+y²）
2013 * 2011y & # 178; ≤ 1, - 1 / (2013 * 2011) ≤ y ≤ 1 / radical (2013 * 2011)
M = 1 + 1 / radical (2013 * 2011), M = 1-1 / (2013 * 2011)
M+m=2

F (x) = (SiNx cosx) the maximum of SiNx

f(x)=sin^2x-cosxsinx
f(x)=(1-cos2x-sin2x)/2
f(x)=(1-(cos2x+sin2x))/2
F (x) = (1-2 ^ (1 / 2) sin (2x + 45 degrees)) / 2
max=(1+2^(1/2)*1)/2
=(1+2^(1/2))/2